$f(x, y) = \left( 16x^3y^2 + \dfrac{1}{2}, 8x^4y \right)$ Find $F$ such that $f = \nabla F$. $F(x, y) =$ $ + \, C$
Solution: We know that $\nabla F = f$. Therefore: $\begin{aligned} F_x &= 16x^3y^2 + \dfrac{1}{2} \\ \\ F_y &= 8x^4y \end{aligned}$ Let's integrate these two equations. Instead of getting a constant at the end of each integral, we'll get a function of the variable with respect to which we didn't integrate. [Example] $\begin{aligned} F &= \int F_x \, dx \\ \\ &= \int 16x^3y^2 + \dfrac{1}{2} \, dx \\ \\ &= 4x^4y^2 + \dfrac{x}{2} + H(y) \\ \\ F &= \int F_y \, dy \\ \\ &= \int 8x^4y \, dy \\ \\ &= 4x^4y^2 + G(x) \end{aligned}$ Now we can set both ways of writing $F$ equal to find $G$ and $H$. $4x^4y^2 + \dfrac{x}{2} + H(y) = 4x^4y^2 + G(x)$ Therefore: $\begin{aligned} G(x) &= \dfrac{x}{2} + C_1 \\ \\ H(y) &= C_2 \end{aligned}$ We can write $C_1$ and $C_2$ as a single arbitrary constant $C$ in the final version of $F$. Putting everything together: $F(x, y) = 4x^4y^2 + \dfrac{x}{2} + C$